Can an F-vector space be defined in multiple ways with multiple dimensions? The same question is found at Dimension of a vector space with addition and scalar multiplication, but it seems the question was never answered and the comments seem to go way off topic. The next paragraph clarifies some points on this page.
The definition of an $F$-vector space $V$ requires that $V$ be a nonempty set such that vector addition $+$ and scalar multiplication $\cdot$ are defined in a certain way. My understanding is that if we change either $\cdot$ or $+$ in a valid way, then we still call $V$ an $F$-vector space. Of course, when referring to certain vector spaces, such as $\mathbb{R}$, there is an agreed upon definition for these operations, such as when viewing $\mathbb{R}$ as a $\mathbb{Q}$-vector space.
My questions: (here the field $F$ stays fixed: as a set and its operations)
(a) If I have an $F$-vector space called $V$ with given $\cdot$ and $+$, is it possible to define a different $\cdot$ and $+$ (whether one changes or both) so that $V$ is still an $F$-vector space AND so that the dimension changes?
(b) If this is possible, should this have any impact on how we define the dimension of $V$ as an $F$-vector space?
(c) When researching general vector spaces, should we care if there are multiple ways to make $V$ an $F$-vector space? I.e. will assuming that "when referring to vector spaces, there is a fixed $\cdot$ and $+$ for the remainder of the context" always never poses a problem?
 A: I'm just going to sort of repeat what was said in the question you linked, since I think it does answer the question.
The thing to understand is that if you take a vector space $V$, and you allow yourself to change the addition and the action of the scalars, then the only thing that remains... is a set. And a set does not have a dimension. So yes you can change the dimension if you change the operations (unless you work over a finite field, in which case the dimension is actually encoded in the cardinality).
Explicitly, suppose $(V,+,\cdot)$ has finite dimension $n\neq 0$, and $(W,+,\cdot)$ has dimension $m\neq 0$. If the base field is infinite you always find a bijection $f: V\to W$. Then we can define a second vector space structure on $V$ by
$$x \bar{+} y = f^{-1}(f(x)+f(y))$$
and
$$\lambda \bar{\cdot} x = f^{-1}(\lambda\cdot f(x)).$$
This defines a vector space $(V,\bar{+},\bar{\cdot})$ over $F$ which is isomorphic to $W$, so it has dimension $m$.
To answer your question (c), in general, in practice a given set does not have multiple natural vector space structures. I am not saying it cannot happen, but no reasonable basic example comes to my mind. So you can feel safe saying "the vector space $V$" without any confusion arising. Of course if you happen to work in a case where there can be a confusion, then you should make things clear.
